What is taylor f(x)=2^x1 ?

The answer to taylor f(x)=2^x1 is 1+ln(2)x+(ln^2(2))/2 x^2+(ln^3(2))/6 x^3+(ln^4(2))/(24)x^4+\ldots

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taylor f(x)=2^x1

taylor

f (x) = 2^{x} 1

1 + ln (2) x + \frac{ln ^{2} ( 2 )}{2} x^{2} + \frac{ln ^{3} ( 2 )}{6} x^{3} + \frac{ln ^{4} ( 2 )}{24} x^{4} + \dots

Solution steps

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( 2 ^{x} \cdot 1 ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 2 ^{x} \cdot 1 ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 2 ^{x} \cdot 1 ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{ln ( 2 )}{1 !} x + \frac{ln ^{2} ( 2 )}{2 !} x^{2} + \frac{ln ^{3} ( 2 )}{3 !} x^{3} + \frac{ln ^{4} ( 2 )}{4 !} x^{4} + \dots

Refine

= 1 + ln (2) x + \frac{ln ^{2} ( 2 )}{2} x^{2} + \frac{ln ^{3} ( 2 )}{6} x^{3} + \frac{ln ^{4} ( 2 )}{24} x^{4} + \dots

\frac{d y}{d t} - 2 t y = 3 t^{2} e^{t^{2}}

\int_{0}^{π} cos^{2} (t) d t

f (x) = x^{e^{x}}

\int (x^{2} y^{3} - \frac{1}{1 + 9 x ^{2}}) d x

\int \frac{14 x - 13}{x ^{2} - 4 x - 21} d x

What is taylor f(x)=2^x1 ?
The answer to taylor f(x)=2^x1 is 1+ln(2)x+(ln^2(2))/2 x^2+(ln^3(2))/6 x^3+(ln^4(2))/(24)x^4+\ldots